CODEWITHSUNDEEP
arraystypescriptobject
user1938143
user1938143

Reputation: 1184

How to get different element by comparing two array object in typescript

I have a problem to get the diffrent element from 2 array object. my example is just like this:

array1= [{id: 1, a: "a", b: "b"}, {id: 2, c: "c", d: "d"}, {id: 3, e: "e", f: "f"}];
array2 = [{c: "c", d: "d"}];

what I expected is:

the output Result is just like this:

this id in every object is not relevant.

result =[{a: "a", b: "b"},{e: "e", f: "f"}];

I have try to use filter or find in typescript, but is was not working.

any solutions??

Upvotes: 0

Views: 58

Answers (2)

Anatoly
Anatoly

Reputation: 22758

Using lodash package and its isEqual method it may look like this:

const _ = require('lodash');

array1= [{id: 1, a: "a", b: "b"}, {id: 2, c: "c", d: "d"}, {id: 3, e: "e", f: "f"}];
array2 = [{c: "c", d: "d"}];

const results = array1.filter(item => !array2.some(item2 => _.isEqual(_.omit(item, ['id']), _.omit(item2, ['id']))))

That way you can compare objects with more that one level of props

Upvotes: 2

wangdev87
wangdev87

Reputation: 8751

array1= [{a: "a", b: "b"}, {c: "c", d: "d"}, {e: "e", f: "f"}];
array2 = [{c: "c", d: "d"}];

const result = array1.filter(subary => {
  var flag = false;

  if (Object.keys(subary).length == Object.keys(array2[0]).length) {
    for (key in subary) {
      if (subary[key] == array2[0][key]) {
        continue;
      } else {
        flag = true;
        break;
      }
    }
  } else {
    flag = true;
  }
  return flag;
})

console.log(result);

Upvotes: 0

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